Engineering task and calculation objective
This module provides the fundamental heat exchanger calculation methods of the VDI Heat Atlas, chapter C1 (12th edition, 2019). It links the heat balance of the two streams with the rate equation Q = k·A·ΔTm and answers the two standard questions of practice: what transfer area does a required duty need (sizing/design calculation), and what outlet temperatures does an existing apparatus deliver (rating calculation)?
The classical methods are available for this: the log mean temperature difference (LMTD) with correction factor F for the actual flow arrangement, and the dimensionless operating characteristic (NTU method) with the number of transfer units and the heat capacity rate ratio. Pure counter-current and co-current flow are covered, as are shell-and-tube exchangers with multiple tube-side and shell-side passes, baffles, cross-flow configurations, plate heat exchangers and stirred vessels with heating or cooling equipment.
The module is needed in process engineering design as well as in operation: for the preliminary sizing of new equipment, for rating existing heat exchangers after a change in operating conditions, or for assessing performance losses due to fouling. The overall heat transfer coefficient k itself is determined in the companion chapter C2.
Standard and calculation basis: VDI-Wärmeatlas, 12. Auflage 2019
Calculation workflow
- Set up the heat balance: The transferred heat flow is balanced from the mass flows, specific heat capacities and inlet/outlet temperatures of both streams. In a design calculation, three of the four temperatures are given and the fourth follows from the balance; in a rating calculation, both outlet temperatures are unknown.
- Define the flow arrangement: Counter-current, co-current, cross-flow or mixed forms (e.g. a shell-and-tube exchanger with one shell-side and two tube-side passes) determine how effectively the driving temperature difference is exploited. For each flow arrangement, the VDI Heat Atlas provides the correction factor F or the corresponding operating characteristic.
- Determine the mean temperature difference or NTU: In the LMTD method, the log mean temperature difference is formed from the temperature differences at the two ends of the apparatus and corrected with F. In the NTU method, the heat capacity rate ratio and the number of transfer units NTU = k·A/Ẇ are formed, and from them the dimensionless temperature change of the streams is calculated.
- Calculate area or duty: Design: from Q, k and the corrected mean temperature difference, the required transfer area A follows. Rating: from the existing area and k, the operating characteristic yields the outlet temperatures and the actual duty of the apparatus.
- Check the result: Plausibility checks complete the calculation: no temperature cross for the chosen flow arrangement, correction factor F not too small (otherwise choose a different configuration), and sufficient margin for fouling via the fouling resistances included in k.
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| t_1' | ϑ'1 | °C |
| t_2' | ϑ'2 | °C |
| t_1'' | ϑ'1 | °C |
| t_2'' | ϑ''2 | °C |
| M_1 | Ṁ1 | kg/s |
| M_2 | Ṁ2 | kg/s |
| c_pm1 | cpm1 | J/(kg·K) |
| c_pm2 | cpm2 | J/(kg·K) |
| W_1 | Ẇ1 | W/K |
| W_2 | Ẇ2 | W/K |
| NTU ratio | ε | - |
| Tube rows | n | - |
| Temperature of the heating/cooling medium (inlet) | ϑ' | °C |
| Temperature of the medium to be heated/cooled | ϑs | °C |
| Mass flow (heating/cooling medium) | Ṁ | kg/s |
| Specific heat capacity (heating/cooling medium) | cpm | J/(kg·K) |
| Heat capacity rate (heating/cooling medium) | Ẇ | W/K |
| (k∙A) of internal tube (heating/cooling medium) | (k∙A)i | W/K |
| (k∙A) of external tube (heating/cooling medium) | (k∙A)a | W/K |
| P | P = f(NTU) = | - |
| Passes | n | - |
| Apparates | Apparates | - |
| Bauform | Bauform | - |
Calculated results
| Quantity | Symbol | Unit |
|---|---|---|
| k | k = A = (k∙A) = | W/(m²·K) |
| A | k = A = (k∙A) = | m² |
| A) | k = A = (k∙A) = | W/K |
| Q | Q̇ = F = Θ = | W |
| Theta | Q̇ = F = Θ = | - |
| F | Q̇ = F = Θ = | - |
| Temperature of the tube flow between the passes | ϑ2z' | °C |
| Temperature of the tube flow between the passes | ϑ2z'' | °C |
| NTU | NTU = P = ϑ'' = | - |
| P | P = f(NTU) = | - |
| t'' | NTU = P = ϑ'' = | °C |
| F_n | Q̇ = F = Fn = | - |
Worked example
An oil cooler is to cool hot water from 90 °C to 50 °C in pure counter-current flow; as the cooling medium, water at 2.0 kg/s is heated from 20 °C to 40 °C. The overall heat transfer coefficient is taken as k = 800 W/(m²·K). Find the heat flow, the log mean temperature difference and the required transfer area — a worked example of a heat exchanger sizing calculation per the VDI Heat Atlas.
Given values
| Inlet/outlet hot stream | 90 °C / 50 °C |
| Inlet/outlet cold stream | 20 °C / 40 °C |
| Mass flow of cold stream | 2.0 kg/s |
| Specific heat capacity of water | 4.19 kJ/(kg·K) |
| Overall heat transfer coefficient k | 800 W/(m²·K) |
| Flow arrangement | pure counter-current (F = 1) |
Solution
Heat flow from the balance of the cold stream
Q = ṁ · cp · (Tout − Tin) = 2.0 kg/s · 4.19 kJ/(kg·K) · (40 − 20) K = 167.6 kW
Log mean temperature difference
In counter-current flow, at the two ends of the apparatus: ΔT1 = 90 − 40 = 50 K and ΔT2 = 50 − 20 = 30 K.
ΔTm = (ΔT1 − ΔT2) / ln(ΔT1/ΔT2) = (50 − 30) / ln(50/30) = 20 / 0.5108 = 39.15 K
Required transfer area
A = Q / (k · ΔTm) = 167,600 W / (800 W/(m²·K) · 39.15 K) = 5.35 m²
In practice, the area is executed with a fouling margin, unless the fouling resistances are already included in the k value.
Result
| Heat flow Q | 167.6 kW |
| Log mean temperature difference | 39.15 K |
| Required area A | 5.35 m² |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
When is the NTU method preferable to the LMTD method?
When rating an existing apparatus, both outlet temperatures are unknown — the LMTD method would then have to iterate, whereas the NTU method delivers the outlet temperatures directly from NTU and the heat capacity rate ratio. For design with known temperatures, on the other hand, the LMTD method is the shorter route. Both methods are mathematically equivalent.
What is a temperature cross and when is it permissible?
A temperature cross occurs when the outlet temperature of the cold stream lies above the outlet temperature of the hot stream. In pure counter-current flow this is perfectly possible; in co-current flow it is fundamentally impossible. In mixed forms such as the 1-2 shell-and-tube exchanger, a sharply dropping correction factor F indicates that the desired cross cannot be achieved economically with this configuration — then several units in series or a more counter-current-like arrangement are needed.
What correction factor F is still considered acceptable?
As a rule of thumb, F should not fall below about 0.75 to 0.8. Smaller values mean not only poor area utilization but also a steep gradient of the F curve: small deviations of the operating temperatures then lead to large changes in duty, and the design becomes non-robust. In such cases, a different flow arrangement or a series configuration is the better solution.
How are special cases such as evaporation or condensation handled?
For an isothermal phase change of one stream, its heat capacity rate is formally infinite; the heat capacity rate ratio becomes zero, and all flow arrangements behave identically — the operating characteristic simplifies accordingly. For partial condensation or temperature-dependent fluid properties, the apparatus must be calculated section by section (cell method).